If ABC and DEF are both 3-digit numbers such that A, B, C, D, E and F are distinct non-zero digits such that ABC + DEF = 1111, then what is the value of A + B + C + D + E + F?
Answer & explanation
Answer: (d) 31
Add column by column. With non-zero digits the units must give C + F = 11 with a carry, then B + E + 1 = 11 and A + D + 1 = 11. So the six digits add up to 11 + 10 + 10 = 31.
- Units column: C + F must end in 1. C + F = 1 is impossible with non-zero digits, so C + F = 11 and 1 is carried.
- Tens column: B + E + 1 must end in 1, so B + E + 1 = 11, i.e. B + E = 10, and 1 is carried.
- Hundreds column: A + D + 1 must give 11 (the leading '11' of 1111), so A + D = 10.
- A + B + C + D + E + F = 10 + 10 + 11 = 31.
- Check: 124 + 987 = 1111, and 1 + 2 + 4 + 9 + 8 + 7 = 31.
Remember · In digit-addition puzzles, work column by column from the units, tracking carries; the digit sum then follows without finding the digits.
Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·